Question

What is the volume occupied by \[25.2g\] of \[C{O_2}\] at \[0.84atm\] and \[{25^ \circ }C\]?

Answer 1

Hint: We need to know the relationship between Pressure, Volume, and temperature and accordingly find the volume. This relationship is given by the Ideal Gas Law. It is a combination of various gas laws such as Boyle’s Law, Charles’s Law, Gay Lussac’s Law and Avogadro’s Law.
We can write the equation for the Ideal Gas Law is
$PV = nRT$,
Where $P$ is the pressure ,$V$ is the Volume, $n$ is the number of moles ,$R$ is the Gas Constant and $T$ is the temperature.

Complete answer:
We have to remember that an ideal gas is one whose particles have negligible volume, are equally sized and do not possess intermolecular forces, have a random motion and perfect elastic collision with no energy loss. We need to remember that an ideal gas equation is given as \[PV = nRT\]. An ideal gas strictly obeys the equation \[PV = nRT\] where Pressure ($P$), Volume (\[V\]), number of moles of gas\[\left( n \right)\]and Temperature ($T$) are taken into consideration.
We know \[PV = nRT\]
\[V = \dfrac{{nRT}}{P}\] where, \[n = \dfrac{{25.2g}}{{44.01g/mol}}\](weight/molecular weight of \[C{O_2}\])
 \[R = 0.0821\dfrac{{L.atm}}{{K.mol}}\]
\[T = {25^ \circ }C = 298K\]
\[P = 0.84atm\]
Now we can substitute the known values in known formula we get,
Therefore, \[V = \dfrac{{\dfrac{{25.2g}}{{44.01g/mol}} \times 0.0821\dfrac{{L.atm}}{{K.mol}} \times 298K}}{{0.84atm}}\]
On simplification we get,
\[V = 17L\]
Hence, the volume occupied by \[25.2g\] of \[C{O_2}\] at \[0.84atm\] and \[{25^ \circ }C\] is \[17L\].

Note:
Note that all of the thermodynamic properties of an ideal gas are summed up in its equation of state, which determines the relationship between its pressure, volume, and temperature. The ideal gas law is one of the simplest equations of state. Although reasonably accurate for gases at low pressures and high temperatures, it becomes increasingly inaccurate at higher pressures and lower temperatures.